Question

Predict whether reactions should occur between aqueous solutions of the following compounds. If so, write balanced molecular and net ionic equations for the reactions. Include physical states. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) and \(\mathrm{CrCl}_{3}\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(\mathrm{HCl}\) (c) \(\mathrm{FeCl}_{3}\) and \(\mathrm{NaOH}\) (d) \(\mathrm{Ca}\left(\mathrm{HCO}_{2}\right)_{2}\) and \(\mathrm{HNO}_{5}\)

Short answer

a) Yes, reaction. 2(NH_{4})_{2}CO_{3(aq)}+ 3CrCl_{3(aq)} \rightarrow Cr_{2}(CO_{3})_{3(s)} + 6NH_{4}Cl_{(aq)}, net ionic: 2NH_{4}^{+} + CO_{3}^{2-} \rightarrow 2NH_{4}^{+} + CO_{3}^{2-}\nb) Yes, reaction. Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_{2}O_{(l)}, net ionic: Ba^{2+} + 2OH^{-} + 2H^{+} + 2Cl^{-} \rightarrow Ba^{2+} + 2Cl^{-} + 2H_{2}O\nc) Yes, reaction. FeCl_{3(aq)} + 3NaOH_{(aq)} \rightarrow Fe(OH)_{3(s)} + 3NaCl_{(aq)}, net ionic: Fe^{3+} + 3OH^{-} \rightarrow Fe(OH)_{3}\nd) No reaction.

Step-by-step solution

Solubility Rules

Consult a chart of solubility rules to predict whether or not the compounds will react. This will determine if a reaction takes place.

Balance Molecular Equations

(a) An aqueous solution of \((NH_4)_2CO_3\) will react with \(CrCl_3\) to form \(Cr_2(CO_3)_3\) and \(NH_4Cl\). After balancing, the balanced molecular equation is: \[2(NH_{4})_{2}CO_{3(aq)}+ 3CrCl_{3(aq)} \rightarrow Cr_{2}(CO_{3})_{3(s)} + 6NH_{4}Cl_{(aq)}\] (b) An aqueous solution of \(Ba(OH)_2\) will react with \(HCl\) to form \(BaCl_2\) and \(H_2O\). The balanced molecular equation is: \[Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_{2}O_{(l)}\] (c) An aqueous solution of \(FeCl_3\) will react with \(NaOH\) to form \(Fe(OH)_3\), a solid, and \(NaCl\), which remains in solution. The balanced molecular equation is: \[FeCl_{3(aq)} + 3NaOH_{(aq)} \rightarrow Fe(OH)_{3(s)} + 3NaCl_{(aq)}\] (d) An aqueous solution of \(Ca(HCO_2)_2\) with \(HNO_5\) will not react because all components are soluble.

Write Net Ionic Equations

(a) For the reaction between \((NH_4)_2CO_3\) and \(CrCl_3\), the net ionic equation, after cancellation of the spectator ions, is: \[2NH_{4}^{+} + CO_{3}^{2-} \rightarrow 2NH_{4}^{+} + CO_{3}^{2-}\] (b) For the reaction between \(Ba(OH)_2\) and \(HCl\), the net ionic equation, after cancellation of the spectator ions, is: \[Ba^{2+} + 2OH^{-} + 2H^{+} + 2Cl^{-} \rightarrow Ba^{2+} + 2Cl^{-} + 2H_{2}O\] (c) For the reaction between \(FeCl_3\) and \(NaOH\), the net ionic equation, after cancellation of the spectator ions, is: \[Fe^{3+} + 3OH^{-} \rightarrow Fe(OH)_{3}\] (d) There is no reaction between \(Ca(HCO_2)_2\) and \(HNO_5\), so there is no net ionic equation.

Key concepts

Solubility Rules

Understanding solubility rules is a foundational skill in predicting chemical reactions, particularly in aqueous solutions. Solubility rules help determine whether a compound is soluble or insoluble in water.
For students grappling with such concepts, it's important to recognize that solubility rules are not arbitrary - they are based on the tendencies of various ions in solution to either remain dispersed or to come together and form a precipitate. For example, nitrates (NO_3^-) are generally soluble, while carbonates (CO_3^{2-}) often are not, unless they're paired with certain cations like NH_4^+ or a group 1 metal cation.
In an exercise like this, you'd consult a solubility chart to predict if compounds like (NH_4)_2CO_3 will react with CrCl_3. According to the rules, carbonate compounds are commonly insoluble, except when combined with ammonium or alkaline metals, suggesting that a reaction that generates a carbonate might result in a precipitate. This insight would lead to the prediction of a solid Cr_2(CO_3)_3 forming in the exercise's part (a).

• Ammonium (NH_4^+) salts are usually soluble.
• Nitrates (NO_3^-) and most chlorides (Cl^-) are soluble, with some exceptions.
• Hydroxides (OH^-) are usually insoluble, except those of Group 1 metals and Ba(OH)_2.
• Sulfates (SO_4^{2-}) are generally soluble, with exceptions like calcium sulfate.

By applying these rules, we can reasonably predict which compounds in a reaction will form a separate phase, aiding in the balance of subsequent molecular equations.

Balanced Molecular Equations

The balanced molecular equation is the stoichiometric representation of a chemical reaction where the number of atoms for each element, and the total charge, is the same on both sides of the equation.
For students, balancing these equations is often an exercise in understanding the conservation of mass and charge. It's crucial to ensure that for every reactant atom there is an equivalent product atom. In example (a) we see 2(NH_4)_2CO_3 and 3CrCl_3 reacting to form Cr_2(CO_3)_3 and NH_4Cl in balanced form, affirming that both the reactants and products have equal numbers of each type of atom, satisfying the law of conservation of mass.
When attempting to balance a molecular equation, starting with the more complex molecules first often simplifies the process. Here, balancing Cr_2(CO_3)_3 before the simpler NH_4Cl can make the task easier, as illustrated in the exercise's step 2.

• Count atoms of each element before and after the reaction.
• Begin with substances that contain the greatest variety of atoms.
• Adjust coefficients, not subscripts, to balance equations.

Balancing equations lays the groundwork for writing net ionic equations, as it establishes the overall stoichiometry of the reaction involved.

Net Ionic Equations

Moving beyond the complete formulas of a reaction's components, net ionic equations break down to the essence of chemical change - focusing solely on the species that undergo a change.
Often, students find the leap from molecular to net ionic equations challenging because it requires a clear understanding of which ions are actively participating in the reaction. The key here is to identify and omit the spectator ions that remain unchanged on both sides of the reaction. These are the ions that appear in exactly the same form on both the reactant and product sides of the balanced molecular equation.
For example, in the given exercise (b), the Ba^{2+} and OH^- form water with H^+, while Cl^- ions do not participate and are therefore removed from the net ionic equation.
Here are some steps to guide students in writing net ionic equations:

• Start with the balanced molecular equation.
• Break down all aqueous compounds into their respective ions.
• Identify and remove the spectator ions that do not change during the reaction.
• Ensure the remaining equation is balanced in terms of both mass and charge.

In some cases, such as exercise (d), you may find that all ions are spectators, which indicates no reaction occurs. Sharing this distinction helps clarify why some mixtures simply don't react and why there are no net ionic equations for those scenarios.